Геометрическая прогрессия падающего мяча

Problem

A ball is dropped from $16$ feet and each bounce reaches $\dfrac{3}{4}$ of the previous height; find the height after the fifth bounce and the total vertical distance traveled before the ball comes to rest.

Step 1: Write the bounce heights as a geometric sequence

The rebound heights form a geometric sequence with starting value $a_0 = 16$ and ratio $r = \dfrac{3}{4}$. For the fifth bounce,

$$a_5 = 16\left(\dfrac{3}{4}\right)^5.$$

Evaluating gives

$$\left(\dfrac{3}{4}\right)^5 = \dfrac{243}{1024}, \qquad 16 \cdot \dfrac{243}{1024} = \dfrac{243}{64} \approx 3.80.$$

So after the fifth bounce, the ball rises to about $3.80$ feet.

Step 2: Sum the repeated bounce distances

The ball first falls $16$ feet. After that, each bounce contributes an up-and-down distance, so the rebound heights are doubled. The bounce heights are

$$12,\ 9,\ 6.75,\ \dots$$

This is an infinite geometric series with first term $a = 12$ and ratio $r = \dfrac{3}{4}$. Its sum is

$$S = \dfrac{12}{1 - \dfrac{3}{4}} = \dfrac{12}{\dfrac{1}{4}} = 48.$$

So the total distance is

$$16 + 2(48) = 112.$$

Answer

The fifth-bounce height is $\dfrac{243}{64} \approx 3.80$ feet, and the total distance traveled is $112$ feet.